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The search() method returns the index of the target x: search(a, 44) returns 2, because a[2] = 44; search(a, 77) returns 5, because a[5] = 77. The method returns 1 when the target is not in the array: search(a, 50) returns 1, because 50 is not in the array.

To obtain a Butterworth filter with a cutofl' frequency R, = 0.3666, we perform the low-pass-to-low-pass transformation r

APPENDIX]

FILTER DESIGN This gives Hob) =

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The sequential search is correct This means that it works The following argument is a proof of that fact If n = 0, then the sequence is empty and the loop does not execute at all Only step 4 executes, immediately returning 1 This satisfies the postconditions: x cannot equal any of the elements because there aren t any If n = 1, then the loop iterates only once, with i = 0 On that iteration, either s0 = x or s0 x If s0 = x, then 0 is returned and the postcondition is satisfied If s0 x, then the loop terminates, step 4 executes, and 1 is returned, and that satisfies the postcondition because the single element of the sequence is not equal to x Suppose n > 1.

(0.3666)'

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b Let m = x and n = x Then by definition, m x < m + 1 and n 1 < x n Then x 1 < m and n < x + 1 Thus x 1 < m x n < x + 1 x < x + 1 merely summarize those in b above c The inequalities x 1 < x x d Let n Z such that n x < n + 1, and let A = {m Z | m x} Then n A and x = maxA, x Now if n < x , then n + 1 x (since both n and x are integers) But n + 1 so n x , by hypothesis Therefore, n = x The proof of the second part is analogous e Assume that x Z (ie, x is an integer).

Finally, we apply the bilinear transformation I - z-' I 2-'

We want to apply the First Principle of Mathematical Induction to deduce that the loop invariant must be true on every iteration of the loop (See page 321) That requires the verification of the invariant on the first iteration and the deduction of the invariant on iteration i from the corresponding invariant on iteration i 1 On the first iteration of the loop, i = 0, and the loop invariant in step 2 is true vacuously because the subsequence {s0..si 1} is empty Then in step 3, either s0 = x or s0 x If s0 = x, then 0 is returned and the postcondition is satisfied If s0 x, then the loop continues on to a second iteration Then i = 1, and the loop invariant in step 2 is again true because the subsequence {s0..si 1} = {s0} and s0 x.

which yields the digital filter H (z) = 0 . 0 0 9 0 ~ - ~ 0.0120~--' 0 . 0 0 9 0 ~ - ~ 0 . 0 0 3 6 ~ - ~ 0.0006~-' 0.0006 0.0036z-' I - 3 . 2 9 4 2 ~ ' 4.8985zr2 - 4.0857zr3 1 . 9 9 3 2 ~ - ~0 . 5 3 5 3 ~ -+ 0 . 0 6 1 5 ~ - ~ ~

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